I'm trying to calculate the limit
$$ L=\lim_{n \to \infty} \left (\ln^2(n)\ln \left (\frac{\ln(\frac{n}{t})}{\ln(\frac{n}{t+1})}\right )-\ln(n)\ln \left (1+\frac{1}{t} \right) \right),$$ where $t$ is a fixed positive integer, but have so far been unsuccessful. I have reason to suspect that $$L=\frac{1}{2}\ln \left (1+\frac{1}{t}\right )\ln \Big(t(t+1) \Big)=\frac{1}{2} \Big(\ln^2(t+1)-\ln^2(t) \Big) \hspace{15mm} (\ast)$$ because (1). Wolfram alpha sais so (without providing any step by step solution ) and (2). It seems to be correct numerically.
Let $$F(n):=\ln^2(n)\ln \left (\frac{\ln(\frac{n}{t})}{\ln(\frac{n}{t+1})}\right )-\ln(n)\ln \left (1+\frac{1}{t} \right).$$ For simplicity, we may take $n=e^m$, so that $$F^*(m)=m^2\ln \left (\frac{m-\ln(t)}{m-\ln(t+1)}\right )-m\ln \left (1+\frac{1}{t} \right)$$ and \begin{equation}\exp(F^*)=\frac{\left(\frac{m-\ln(t)}{m-\ln(t+1)} \right )^{m^2}}{(1+\frac{1}{t})^{m}}= \left(\frac{t}{t+1} \right )^m\left(\frac{m-\ln(t)}{m-\ln(t+1)} \right )^{m^2}\end{equation} I do not know how to calculate the limit of this as $m \to \infty$. One could substitute $a=\frac{t}{t+1}$ and then set $b=\ln(a)$. The result is (any of) these expression: $$a^{m} \left (1+\frac{\ln(a)}{x-\ln(t/a)} \right )^{m^2} =e^{bm} \left(1+\frac{b}{x+b-\ln(t)}\right )^{m^2}=e^{bm} \left(1+\frac{b}{x-\ln(t+1)}\right )^{m^2}.$$ If we pick, say the rightmost expression above and take the logarithm, the problem is reduced to calculating the limit $$\lim \limits_{m \to \infty} \left(bm+m^2\ln \left (1+\frac{b}{m-\ln \left( t+1 \right ) } \right ) \right )$$ $$=\lim \limits_{m \to \infty} m\left(b+\ln \left (\left[1+\frac{b}{m-\ln \left( t+1 \right ) } \right )\right]^{m} \right)$$ $$=\lim \limits_{m \to \infty} m \left( b+ \ln \left ( \left[ 1+\frac{b}{m-\ln(t+1) } \right]^{m-\ln(t+1)} \left[ 1+\frac{b}{m-\ln(t+1) } \right]^{\ln(t+1)} \right) \right)$$ which does at least have the appearance of a simpler problem. Then, setting $m=\ln(t+1)+x$ we must calculate the limit of $$\Big(\ln(t+1)+x\Big) \left(b+\ln \left(\left[1+\frac{b}{x} \right]^{x} \right)+\ln(t+1)\ln \left(1+\frac{b}{x} \right ) \right)$$ as $x$ goes to infinity. (The plan in this calculation is to use $e^b=\lim_{x \to \infty}(1+b/x)^x$) as some point. I have also tried to calculate the limit $(\ast)$ using L'Hopitals rule, but to no avail. Does anyone see the golden step i am missing ? : )
Let $a=\ln(t)$, $b=\ln(t+1)$ and $x=1/\ln(n)$. Then $x\to 0$ as $n\to +\infty$ and $$\ln^2(n)\ln \left (\frac{\ln(\frac{n}{t})}{\ln(\frac{n}{t+1})}\right )-\ln(n)\ln \left (1+\frac{1}{t} \right)=\frac{\ln(1-ax)-\ln(1-bx)-(b-a)x}{x^2}.$$ Now by using the Taylor series $$\ln(1+y)=y-\frac{y^2}{2}+O(y^3),$$ we find that the above LHS is $$\frac{b^2-a^2}{2}+O(x).$$ Therefore the limit $L$ is equal to $$\frac{b^2-a^2}{2}=\frac{\ln^2(t+1)-\ln^2(t)}{2}$$ as you expected.